Find the shortest distance from the point $P=(5,4,-6)$ to the line
$$(x,y,z)=(-7t,-2t,t)$$
So, I'm trying to find the shortest distance and here is my setup:
$$(x,y,z)=(0,0,0) + t(-7,-2,1)$$
Line equation: $-7x – 2y + z = 0$.
Using my point given I have $$\frac{-7(5) – 2(4) + (-6)}{\sqrt{(-7)^2 + (-2)^2 + 1^2} }$$
I get $6.395$ units.
What am I doing wrong here?
Best Answer
Let the line be defined by $\mathcal{L} := \{ (-7t,-2t,t) : t \in \mathbb{R} \}$. The distance between $\mathcal{L}$ and $P$ is
$$\text{dist} (\mathcal{L},P) := \displaystyle\min_{t \in \mathbb{R}}\sqrt{ (7 t + 5)^2 + (2 t + 4)^2 + (t + 6)^2}$$
Let $t^* := \arg\min \sqrt{ (7 t + 5)^2 + (2 t + 4)^2 + (t + 6)^2}$ be the minimizer. Note that
$$t^* = \displaystyle\arg\min \{ (7 t + 5)^2 + (2 t + 4)^2 + (t + 6)^2\} = -\frac{3}{2}$$
and, therefore, we have that $\text{dist} (\mathcal{L},P) = \sqrt{ (7 t^* + 5)^2 + (2 t^* + 4)^2 + (t^* + 6)^2}$.